Question #306cf

1 Answer
Jul 20, 2017

Answer:

Left side:

#"K"#: #1+#

#"Cl"#: #7+#

#"O"#: #2-#

Right side:

#"K"#: #1+#

#"Cl"#: #1-#

#"O"#: #0#

Explanation:

We're asked to assign an oxidation state to each element on both sides of the equation.

All pure elements have an oxidation state of #0#, so the oxygen on the right side has an oxidation state of #0#.

An important fact worth knowing is that oxygen will almost always have an oxidation state of #2-# when in a compound. (The most common exception to this are superoxides, such as #"KO"_2#).

Thus, oxygen on the left side has an oxidation state of #2-#.

Group #1# metals such as #"Na"# and #"K"# will almost always have a #1+# oxidation state, so the potassium on both sides has an oxidation state of #1+#.

Now, we only have #"Cl"# left.

All neutral compounds have a net oxidation state of #0#, so we use a little simple math to find the oxidation state of chlorine on both sides:

Left side:

#"KClO"_4#: #overbrace(1+)^"K" + overbrace(x)^"Cl" + overbrace(4(2-))^(4color(white)(l)"O atoms") = 0#

#x = color(red)(+7#

The oxidation state of #"Cl"# on the left side is thus #color(red)(7+#.

Right side:

#"KCl"#: #overbrace(1+)^"K" + overbrace(x)^"Cl" = 0#

#x = color(blue)(-1#

The oxidation state of #"Cl"# on the right side is thus #color(blue)(1-#.